Two Earth's satellites move in a common plane along circular obrits. The orbital radius of one satellite is r while that of the other satellite is `r- Deltar("Here"Deltarlt ltr)`

Options

A. Time interval separates the periodic approaches of the satellites to each other over the minimum distance is `(4pir^(5//2))/(3(GM)^(1//2)Deltar)`
B. Time interval separates the periodic approaches of the satellites to each other over the minimum distance is `(2pi^(5//2))/(3(GM)^(1//2)Deltar)`
C. Angular velocity of approach between two setellites is `(3(GM)^(1//2)Deltar)/(r^(5//2))`
D. Angular velocity of approach between two setellites is `(3(GM)^(1//2)Deltar)/(2r^(5//2))`

Explanation

To solve the problem of two Earth satellites moving in circular orbits with different radii, we can follow these steps:

### Step 1: Understand the Problem
We have two satellites:
- Satellite 1 is at radius \( r \).
- Satellite 2 is at radius \( r - \Delta r \), where \( \Delta r \) is very small compared to \( r \).

### Step 2: Calculate Angular Velocities
The angular velocity \( \omega \) of a satellite in orbit can be expressed using the formula:
\[ \omega = \sqrt{\frac{GM}{r^3}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

For Satellite 1 (radius \( r \)):
\[ \omega_1 = \sqrt{\frac{GM}{r^3}} \]

For Satellite 2 (radius \( r - \Delta r \)):
\[ \omega_2 = \sqrt{\frac{GM}{(r - \Delta r)^3}} \]

### Step 3: Find the Angular Velocity of Approach
The angular velocity of approach \( \Delta \omega \) is given by:
\[ \Delta \omega = \omega_1 - \omega_2 \]
Substituting the expressions for \( \omega_1 \) and \( \omega_2 \):
\[ \Delta \omega = \sqrt{\frac{GM}{r^3}} - \sqrt{\frac{GM}{(r - \Delta r)^3}} \]

### Step 4: Simplify the Expression
To simplify \( \Delta \omega \), we can factor out \( \sqrt{GM} \):
\[ \Delta \omega = \sqrt{GM} \left( \frac{1}{r^{3/2}} - \frac{1}{(r - \Delta r)^{3/2}} \right) \]

### Step 5: Use Binomial Expansion
For small \( \Delta r \), we can use the binomial expansion:
\[ (r - \Delta r)^{-3/2} \approx r^{-3/2} \left( 1 + \frac{3\Delta r}{2r} \right) \]
Thus,
\[ \Delta \omega \approx \sqrt{GM} \left( \frac{1}{r^{3/2}} - \left( \frac{1}{r^{3/2}} + \frac{3\Delta r}{2r^{5/2}} \right) \right) \]
This simplifies to:
\[ \Delta \omega \approx -\frac{3\Delta r \sqrt{GM}}{2r^{5/2}} \]

### Step 6: Calculate Time Interval
The time interval \( \Delta t \) between the two satellites can be calculated as:
\[ \Delta t = \frac{2\pi}{\Delta \omega} \]
Substituting our expression for \( \Delta \omega \):
\[ \Delta t = \frac{2\pi}{-\frac{3\Delta r \sqrt{GM}}{2r^{5/2}}} = \frac{4\pi r^{5/2}}{3\Delta r \sqrt{GM}} \]

### Conclusion
The final expressions for angular velocity of approach and time interval have been derived. The correct options based on these calculations would be the ones that match these results.

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